Problem: $g(t) = -7t^{2}+6t-3(f(t))$ $f(x) = x$ $ g(f(0)) = {?} $
Explanation: First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = $ $f(0) = 0$ Now we know that $f(0) = 0$ . Let's solve for $g(f(0))$ , which is $g(0)$ $g(0) = -7(0^{2})+(6)(0)-3(f(0))$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = $ $f(0) = 0$ That means $g(0) = -7(0^{2})+(6)(0)+(-3)(0)$ $g(0) = 0$